1
Experiments with Random Factors
• Previous chapters have considered fixed factors– A specific set of factor levels is chosen for the experiment– Inference confined to those levels– Often quantitative factors are fixed (why?)
• When factor levels are chosen at random from a larger population of potential levels, the factor is random– Inference is about the entire population of levels– Industrial applications include measurement system studies
2
Random Effects Models
• In the fixed effects models, the main interest is the treatment means
• In random effects models, the knowledge about the particular ones investigated is relatively useless, but the variability is the primary concern
• The population of factor levels is of infinite size, or is large enough to be considered infinite
3
Random Effects Models• The usual single factor ANOVA model is
• Now both the error term and the treatment effects are random variables:
• Variance components:
Total variability in the observations is partitioned into1. A component that measures the variability between
the treatments, and2. A component that measures the variability within
treatments
1,2,...,
1, 2,...,ij i ij
i ay
j n
2 2 is (0, ) and is (0, )ij iNID NID 2 2( )ijV y
•The sums of squares SST = SSTreatment + SSE
4
Relevant Hypotheses in the Random Effects (or Components of Variance) Model
• In the fixed effects model we test equality of treatment means
• This is no longer appropriate because the treatments are randomly selected – the individual ones we happen to have are not of
specific interest
– we are interested in the population of treatments
• The appropriate hypotheses are2
0
21
: 0
: 0
H
H
5
Testing Hypotheses - Random Effects Model
• The standard ANOVA partition of the total sum of squares still works; leads to usual ANOVA display
• Form of the hypothesis test depends on the expected mean squares
and• Therefore, the appropriate test statistic is
• Ho is rejected if Fo > F,a-1,N-a
aNSSa
SS
MS
MSF
E
Treatments
E
Treatmentso
1
2)( EMSE 22)( nMSE Treatments
6
Estimating the Variance Components• Use the ANOVA method; equate expected mean squares to
their observed values:
• Potential problems with these estimators– Negative estimates (woops!)– They are moment estimators & don’t have best statistical properties
2 2 2
2
2
ˆ ˆ ˆ and
ˆ
ˆ
E Treatments
Treatments E
E
MS n MS
MS MS
n
MS
7
Random Effects Models• Example 13-1 (pg. 467) – weaving fabric on looms
• Response variable is strength
• Interest focuses on determining if there is difference in strength due to the different looms
• However, the weave room contains many (100s) looms
• Solution – select a (random) sample of the looms, obtain fabric from each
• Consequently, “looms” is a random factor
8
• It looks like a standard single-factor experiment with a = 4 & n = 4
Table 13-1 Strength Data for Example 13-1
9
Minitab Solution (Balanced ANOVA)Factor Type Levels Values
Loom random 4 1 2 3 4
Analysis of Variance for y
Source DF SS MS F P
Loom 3 89.188 29.729 15.68 0.000
Error 12 22.750 1.896
Total 15 111.938
Source Variance Error Expected Mean Square for Each Term
component term (using unrestricted model)
1 Loom 6.958 2 (2) + 4(1)
2 Error 1.896 (2)
10
Random Effects Models• Important use of variance components – isolating
of different sources of variability that affect a product or system
Figure 13-1
11
Confidence Intervals on the Variance Components
• Easy to find a 100(1-)% CI on
• Other confidence interval results are given in the book
• Sometimes the procedures are not simple
2 :
22 2
/ 2, 1 ( / 2),
( ) ( )E E
N a N a
N a MS N a MS
12
Extension to Factorial Treatment Structure
• Two factors, factorial experiment, both factors random (Section 13-2, pg. 490)
• The model parameters are NID random variables• Random effects model
2 2 2 2
2 2 2 2
1,2,...,
( ) 1,2,...,
1, 2,...,
( ) , ( ) , [( ) ] , ( )
( )
ijk i j ij ijk
i j ij ijk
ijk
i a
y j b
k n
V V V V
V y
13
Testing Hypotheses - Random Effects Model
• Once again, the standard ANOVA partition is appropriate
• The sums of squares are calculated as in the fixed effects case
• Relevant hypotheses:
• Form of the test statistics depend on the expected mean squares:
2 2 20 0 0
2 2 21 1 1
: 0 : 0 : 0
: 0 : 0 : 0
H H H
H H H
2 2 20
2 2 20
2 20
2
( )
( )
( )
( )
AA
AB
BB
AB
ABAB
E
E
MSE MS n bn F
MS
MSE MS n an F
MS
MSE MS n F
MS
E MS
14
Estimating the Variance Components – Two Factor Random model
• As before, use the ANOVA method; equate expected mean squares to their observed values:
• Potential problems with these estimators
2
2
2
2
ˆ
ˆ
ˆ
ˆ
A AB
B AB
AB E
E
MS MS
bnMS MS
anMS MS
n
MS
15
Example 13-2 (pg. 492) A Measurement Systems Capability Study
• Gauge capability (or R&R) is of interest
• The gauge is used by an operator to measure a critical dimension on a part
• Repeatability is a measure of the variability due only to the gauge
• Reproducibility is a measure of the variability due to the operator
• This is a two-factor factorial (completely randomized) with both factors (operators, parts) random – a random effects model
16
Table 13-313-2
17
Example 13-2
• The objective is estimating the variance components
• is gauge repeatability
• is reproducibility of the gauge
• The variance components:
222
99.0
14.02
99.071.0
015.0)2)(20(
71.031.1
28.10)2)(3(
71.039.69
2
2
2
2
18
Example 13-2 (pg. 494) Minitab Solution – Using Balanced ANOVA
Source DF SS MS F P
Part 19 1185.425 62.391 87.65 0.000
Operator 2 2.617 1.308 1.84 0.173
Part*Operator 38 27.050 0.712 0.72 0.861
Error 60 59.500 0.992
Total 119 1274.592
Source Variance Error Expected Mean Square for Each Term
component term (using unrestricted model)
1 Part 10.2798 3 (4) + 2(3) + 6(1)
2 Operator 0.0149 3 (4) + 2(3) + 40(2)
3 Part*Operator -0.1399 4 (4) + 2(3)
4 Error 0.9917 (4)
19
Example 13-2 (pg. 494) Minitab Solution – Balanced ANOVA
• There is a large effect of parts (not unexpected)• Small operator effect• No Part – Operator interaction• Negative estimate of the Part – Operator
interaction variance component• Fit a reduced model with the Part – Operator
interaction deleted
yijk = + i + j + ijk
20
Example 13-2 (pg. 494) Minitab Solution – Reduced Model
Source DF SS MS F P
Part 19 1185.425 62.391 70.64 0.000
Operator 2 2.617 1.308 1.48 0.232
Error 98 86.550 0.883
Total 119 1274.592
Source Variance Error Expected Mean Square for Each Term
component term (using unrestricted model)
1 Part 10.2513 3 (3) + 6(1)
2 Operator 0.0106 3 (3) + 40(2)
3 Error 0.8832 (3)
21
Example 13-2 (pg. 495) Minitab Solution – Reduced Model
• The expected mean squares are
• Estimating the variance components
• Estimating gauge capability: 2 2 2ˆ ˆ ˆ
0.88 0.01
0.89
gauge
2
22
22
)(
)(
)(
E
B
A
MSE
anMSE
bnMSE
88.0ˆ
0108.0)2)(20(
88.031.1ˆ
25.10)2)(3(
88.039.62ˆ
2
2
2
E
EB
EA
MS
an
MSMS
bn
MSMS
22
?